NJC 2025 H2 Physics Prelim P3 Ans

2025 SH2 H2 Physics Preliminary Examination Paper 3 9749/03 Section A 1(a) work done per unit mass in bringing a (small test) mass from infinity to that point B1 1(b)(i) correct read off of 𝜙 and x, e.g. B1 − GM R = −6.3 × 107 correct equation and substitution of G, M and R, e.g. B1 𝑀 = 6.3 × 107 × 6.4 × 106 6.67 × 10−11 = 6.0 × 1024 kg (shown) 1(b)(ii)1. attempt to apply Conservation of Energy in any understandable form, e.g. C1 Gain in KE = Loss in GP 1 2 𝑚𝑣2 = 𝑚(𝜙𝑖 − 𝜙𝑓) 𝑣 = √2(0 − (−2.1)) × 107 value of 𝜙𝑓 = −2.1 × 107 J kg−1 (read off when x = 3R) C1 = 6500 m s−1 A1 Comment: the most common mistake was substituting x = 2R. Some candidates thought the mass was in a circular orbit which is the wrong context. 1(b)(ii)2. 𝑎 = 𝐹 𝑚 = 𝐺𝑀𝑚 𝑥2 𝑚 = 𝐺𝑀 (3𝑅)2 = 6.67 × 10−11 × 6.0 × 1024 (3 × 6.4 × 106)2 allow also a = g (ecf awarded if sub x = 2R) C1 = 1.1 m s−2 A1 Comment: similar mistake to the previous part. It is inappropriate to find the gradient on the graph as this is an approximate method. The question asks you to calculate not estimate the acceleration. The more precise method should be chosen. Also, using equations of motion is wrong as the acceleration is not constant here. 1(b)(iii) lower (or more negative) gravitational potential energy due to presence of the Moon M1 hence higher speed A1

2(a)(i) change in momentum of molecule per collision = −mu − mu = −2mu OR = mu − (−mu) = 2mu A1 2(a)(ii) average force (= impulse per collision time between collisions) = 2mu 2x/u M1 = mu2 x (shown) 2(b)(i) W = pV = (7.0×105) [ (20 − 5) ×10−6 ] C1 = 10.5 J A1 Comment: some responses perhaps thought the work done is the area within the cycle. Negative sign should not be applied as this is work done by. 2(b)(ii) heating supplied to gas / J work done on gas / J increase in internal energy of gas / J 36.8 − 10.5 26.3 − 30.0 zero − 30.0 zero 3.7 3.7 third row (C → A) B1 second row (B → C) B1 first row (A → B) – value in red (allow ecf) B1 first row (A → B) – values in blue (allow ecf) B1 Comment: the sign for work done for A → B is sometimes given wrongly 2(b)(iii) useful work done = 10.5 − 3.7 = 6.8 J C1 efficiency = useful work done total energy input × 100% = 6.8 36.8 × 100% = 18% A1 Comment: few got this correct, in particular, the energy input must be the heat supplied to the system. No process can be 100% efficient or more.

3(a) When two or more waves meet at a point, B1 the resultant displacement at that point is equal to the vector sum of the displacements due to the individual waves at that point. B1 Examiner’s comment: The student did not score for this question. 3(b)(i) wavelength = 3.0 × 108 2.5 × 1010 B1 = 0.012 m 3(b)(ii)1. waves are in anti-phase at the sources and have no path difference to reach O M1 waves meet in anti-phase at O and destructive interference occurs A1 Examiner’s comment: Many student failed to state “No path difference when the two waves reach O.” 3(b)(ii)2. path difference = 1 2  = 0.0060 m A1 3(b)(iii) x (= D 𝑎 ) = (0.012)(2.3) 0.18 M1 = 0.15 m A1 3(b)(iv) point B is above or below point O such that distance OB ≈ ½ distance OA B1 O A × B ×

4(a) electric force exerted per unit positive charge placed at that point B1 Examiner’s comment: The student did not score for this question. 4(b)(i) For the electric potential at point P to be zero, the contribution by one sphere must be positive and the other negative, hence opposite signs of charges on spheres M1 electric fields (due to X and Y are not zero and) must be in the same direction A1 hence not zero Examiner’s comment: The student did not score for this question. 4(b)(ii)1. VX = −Q 4π𝜀0x VY = +2Q 4π𝜀0y VX + VY = 0 Q 4π𝜀0x = 2Q 4π𝜀0y B1 y = 2x (shown) 4(b)(ii)2. E = Q 4π𝜀0x2 + 2Q 4π𝜀0(2x) 2 C1 = 3Q 8π𝜀0x2 A1 Examiner’s comment: The student did not score for this question. Many students did not realise that the resultant field is in the same direction. 4(c)(i) horizontal line above zero up to at least 2/3 of the x-axis B1 curves upwards thereafter B1 Examiner’s comment: The student did not score for this question. Many students did not realise that the electric field is constant, i.e a horizontal line between 0 < potential < 1600 V. 4(c)(ii) Both the electron and argon ion travels through the same potential difference and hence loses the same amount of EPE and gains the same amount of KE M1 gain in KE of the electron = gain in KE of the argon ion

1 2 meve2 = 1 2 mAvA2 ve vA = √6.64×10−26 9.11×10−31 = 270 A1 OR gain in KE of the electron = e ∆𝑉 M1 ½ (9.11 x 10-31) v2e = 1.6 x 10-19 (4000 – 2000) -------- (1) gain in KE of the Argon ion = e ∆𝑉 ½ (6.64 x 10-26) v2Ar = 1.6 x 10-19 (2000 – 0) ------------ (2) (1)/(2), ve vA = √6.64×10−26 9.11×10−31 = 270 A1 Examiner’s comment: Most student obtained zero for this question. Many students gave unclear working/no substitution of values in the working/no explanation of why kinetic energy of electron is equal to kinetic of Argon ion. Zero marks were awarded to student as long as student did not give explanation/substitute values inside equation even if their numerical answer is correct. 4(c)(iii) argon encounters more collisions than electron / electron encounters fewer collisions than Argon OR electron is less ionising (than Argon) / argon is more ionising (than electron) and electron loses less energy (than Argon) / argon loses more energy (than electron), hence ratio is larger B1 Examiner’s comment: Most students were able to explain to answer this question. Students must appreciate besides the speed of object, other factors, i.e mass of object, distance travelled by object affects the amount of air resistance.

5(a)(i) Induced emf E = BLv or E is proportional to v where v is the instantaneous velocity of the rod. M1 Graph showed a straight line (passing through the origin) indicating that e.m.f. varies linearly with (directly proportional to) time. Hence the rod’s velocity must be increasing linearly with (directly proportional to) time. Therefore acceleration of the rod is uniform. A1 Comments: 1. Question did not asked students to explain why an emf is induced, hence students should not be wasting time writing about changes in flux linkage / flux cutting, quoting Faraday’s law and explaining why an emf is induced. 2. Those who could recall the expression E = Blv were able to successfully answer the question. 3. Those who tried to explain in terms of rate of flux cutting or rate of change of flux linkage merely paraphrased the question ie “emf varies linearly with time, hence rate of change of flux linkage is constant / increase linearly with time and therefore ac celeration is constant”. 5(a)(ii) kinetic energy is converted to electrical energy (to drive the current around the circuit) / thermal energy in the resistor M1 resulting in the decrease in the kinetic energy of the rod / rod slowing down. A1 hence external work needed to maintain constant kinetic energy / speed Comments: 1. Very few students accounted for the conversion of kinetic energy to electrical energy / thermal energy in the resistor. 2. Majority of students explained in terms of the retarding force exerted by the magnetic field on the induced current / rod causing the rod to slow down. 3. A significant number of students thought that the rod is accelerating and h